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The temperature coefficient of resistivity of material is 0.0004/K. When the temperature of the material is increased by 50 oC , its resistivity increases by 2 x 10-8  ohm -meter. The initial resistivity of the material in ohm-meter is
  • a)
    50 x 10-8
  • b)
    90 x 10-8
  • c)
    100 x 10-8
  • d)
    200 x 10-8
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The temperature coefficient of resistivity of material is 0.0004/K. Wh...
Given:
Temperature coefficient of resistivity (α) = 0.0004/K
Change in temperature (ΔT) = 50°C = 50K
Change in resistivity (Δρ) = 2 × 10^-8 Ω-m

To find: Initial resistivity (ρ)

Formula Used:
Temperature coefficient of resistivity (α) = (1/ρ) × (dρ/dT)

where ρ = resistivity, T = temperature and α = temperature coefficient of resistivity

Calculation:

Let the initial resistivity be ρ0.

As per the given information,
Δρ = ρ - ρ0 = 2 × 10^-8 Ω-m
ΔT = T - T0 = 50 K
α = 0.0004/K

From the formula,
α = (1/ρ) × (dρ/dT)

On integrating both sides, we get,
ln ρ - ln ρ0 = α × ΔT
ln (ρ/ρ0) = α × ΔT
ρ/ρ0 = e^(αΔT)

Substituting the given values,
ρ/ρ0 = e^(0.0004 × 50)
ρ/ρ0 = e^0.02
ρ/ρ0 = 1.02020134014

ρ = ρ0 × 1.02020134014

From the above equation,
ρ = 1.02020134014 ρ0

Substituting the value of Δρ, we get,
2 × 10^-8 = (1.02020134014 ρ0) - ρ0
2 × 10^-8 = 0.02020134014 ρ0
ρ0 = 0.02020134014/2 × 10^-8

ρ0 = 100 × 10^-8 Ω-m

Therefore, the initial resistivity of the material is 100 × 10^-8 Ω-m.

Hence, the correct option is (c) 100 × 10^-8 Ω-m.
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The temperature coefficient of resistivity of material is 0.0004/K. When the temperature of the material is increased by 50 oC , its resistivity increases by 2 x 10-8 ohm -meter. The initial resistivity of the material in ohm-meter isa)50 x 10-8b)90 x 10-8c)100 x 10-8d)200 x 10-8Correct answer is option 'C'. Can you explain this answer?
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